Cycloid Motion with EM Fields
Problem Consider a uniform magnetic field directed in the x-direction, and a uniform electric field directed in the z-direction. A particle with positive charge q and mass m is released from the origin, and initially at rest. Determine the trajectory of the particle over time. : \boldsymbol{E} = \begin{pmatrix} 0 \\ 0 \\ E \end{pmatrix}\ , \quad \boldsymbol{B} = \begin{pmatrix} B \\ 0 \\ 0 \end{pmatrix}\ Solution The magnetic force is calculated using the cross-product : \boldsymbol{v \times B} = \begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\ 0 & \dot y & \dot z \\ B & 0 & 0 \end{vmatrix}\ = \begin{pmatrix} 0 \\ B \dot z \\ -B \dot y \end{pmatrix}\ . Subsequently the Lorentz force, \boldsymbol{F} = q \left(\boldsymbol{E} + \boldsymbol{v \times B} \right) , in vector form is : \begin{pmatrix} m \ddot x \\ m \ddot y \\ m \ddot z \end{pmatrix}\ = \begin{pmatrix} 0 \\ qB \dot z \\ qE - qB \dot y \end{pmatrix}\ Divide the mass of the particle on both sides of the equation. Let \omega = \frac{qB}{m} and \gamma = \frac{qE}{m} , then : \begin{pmatrix} \ddot x \\ \ddot y \\ \ddot z \end{pmatrix}\ = \begin{pmatrix} 0 \\ \omega \dot z \\ \gamma - \omega \dot y \end{pmatrix}\ . To solve this system of coupled differential equations, let \ddot x = \dot v_x, \quad \ddot y = \dot v_y, \quad \ddot z = \dot v_z . Thus : \begin{pmatrix} \dot v_x \\ \dot v_y \\ \dot v_z \end{pmatrix}\ = \begin{pmatrix} 0 \\ \omega v_z \\ \gamma - \omega v_y \end{pmatrix}\ . Substitute the middle equation, into the bottom equation to obtain : \ddot v_y + {\omega}^{2} v_y = \omega \gamma . The above differential equation is non-homogeneous, thus the solution is the sum of the homogeneous solution and particular solution. (1) Homogeneous solution : \ddot v_h + {\omega}^{2} v_h = 0 The characteristic equation is {\lambda}^{2} + {\omega}^{2} = 0 , where \lambda = \pm \omega i . The solution is therefore : v_h = A{e}^{i \omega t} + B{e}^{-i \omega t} or : v_h = C\sin{(\omega t)} + D\cos{(\omega t)} . (2) Particular solution Since \omega \gamma is constant, if v_p = \omega \gamma , then \dot v_p = \ddot v_p = 0 . Then the differential equation : \ddot v_p + {\omega}^{2} v_p = \omega \gamma reduces to : 0 + {\omega}^{2} v_p = \omega \gamma . Hence v_p = \frac{\gamma}{\omega} . Consequently, the solution of the differential equation is : v_y = C\sin{(\omega t)} + D\cos{(\omega t)} + \frac{\gamma}{\omega} . Since the particle is initially released at rest, : v_y = \frac{\gamma}{\omega}\left(1 - \cos{(\omega t)}\right) : v_z = \frac{\gamma}{\omega}\sin{(\omega t)} To obtain the trajectory, : \int v_y dt = \frac{\gamma}{\omega}t - \frac{\gamma} \sin{(\omega t)} + y_0 : \int v_z dt = - \frac{\gamma} \cos{(\omega t)} + z_0 Since the particle is initially released at the origin, : y = \frac{\gamma}{\omega}t - \frac{\gamma} \sin{(\omega t)} : z = \frac{\gamma} \left(1 - \cos{(\omega t)}\right) Replacing \gamma and \omega with its original quantities : y = \frac{E}{B}\left(t - \frac{m}{qB}\sin{\left(\frac{qB}{m} t\right)}\right) : z = \frac{Em}{q{B}^{2}}\left(1 - \cos{\left(\frac{qB}{m} t \right)}\right) . The solution of displacement presents a cycloid trajectory, which is usually given in parametric form as : y = a\left(t - \sin{(t)}\right) : z = a\left(1 - \cos{(t)}\right) . Category:Physics Category:Electromagnetism and Optics Category:Modern puzzles Category:Electromagnetism Category:Electricity and Magnetism Category:Differential Equations